Question: The number $2017$ is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$. What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$

Solution: Note that $2014\equiv -3 \mod2017$. We have for $k\ge1$\[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\]\[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\]\[\equiv (-1)^k\dbinom{k+2}{2} \mod 2017\]Therefore\[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\]This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that\[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{  if n is odd}\]and\[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{  if n is even}\]Obviously, $62$ falls in the second category, so our desired value is\[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{1024}\]